Ch. 6 – Thermal Energy I. Temperature and Energy Transfer (p. 158 – 170) Temperature Thermal...

Ch. 6 – Thermal Energy

I. Temperature and Energy Transfer

(p. 158 – 170) Temperature Thermal Energy Heat Transfer

A. Temperature

Temperature measure of the

average KE of the particles in a sample of matter

Thermometers rely on expansion

B. Temperature Scales

Fahrenheit and Celsius most familiar to you Celsius used often in science – why? Kelvin, based on Celsius and absolute zero Absolute zero = temperature at which an

object’s energy is the lowest Celsius – Kelvin Conversion

K = oC + 273 ? K = 32oC

K = 32oC + 273 = 305K

C. Energy Transfer

Molecules of warmer substance are moving faster than molecules of colder substance

As these substances come in contact with each other, energy is transferred from warm to cold objects because of collisions of particles

Example: ice cube in a warm hand

D. Thermal Energy

Thermal Energy the total energy of the particles in a material KE - movement of particles PE - forces within or between particles due to

position depends on temperature, mass, and type of

substance

D. Thermal Energy

Which beaker of water has more thermal energy? B - same temperature, more mass

200 mL

80ºC

A400 mL

80ºC

B

E. Heat Transfer

Heat thermal energy that flows from

a warmer material to a cooler material Like work, heat is...

measured in joules (J) a transfer of energy

E. Heat Transfer

Why does A feel hot and B feel cold?

80ºC

A

10ºC

B

Heat flows from A to your hand = hot

Heat flows from your hand to B = cold

A B

F. Thermal Energy Transfer

Conduction

Convection

Radiation

G. Conduction

Conduction = movement of heat energy by direct contact Good conductors =

material through which energy can be easily transferred as heat

Ex: metals, saltwater Opposite of conductor

= insulator – slows the transfer of energy as heat

Ex: cork, wood, glass, air

H. Convection

Convection is the movement of gases or liquids from a warmer spot to a cooler spot

A lava lamp displays convection as heated wax expands and rise and cooler wax falls

I. Radiation

Electromagnetic waves - require no matter to travel through; ex: light, IR, UV

Electromagnetic waves transfer energy = radiation

When sunlight hits earth, its radiation is absorbed or reflected

Darker surfaces absorb more of the radiation and lighter surfaces reflect the radiation

J. Thermal Energy Summary

Room is warmed by radiation

K. Heat Transfer

Specific Heat (C) amount of energy

required to raise the temp. of 1 kg of material by 1 degree Kelvin

units: J/(kg·K)or J/(kg·°C) or J/(g·°C)

K. Heat Transfer Which sample will take

longer to heat to 100°C?

50 g Al 50 g Cu

• Al – It has a higher specific heat• Al will also take longer to cool down

K. Heat Transfer

q = m C T

q: heat (J)m: mass (kg)C: specific heat (J/kg·K) T: change in temperature (K or °C)

T = Tf - Ti

– q = heat loss+ q = heat gain

K. Heat Transfer

Calorimeter device used to

measure changes in thermal energy

Coffee cup Calorimeter

in an insulated system:

heat gained = heat lost

K. Heat Transfer

A 32.0-g silver spoon cools from 60.0°C to 20.0°C. How much heat is lost by the spoon?

GIVEN:

m = 32.0 g

Ti = 60.0°C

Tf = 20.0°C

q = ?

C = 235 J/kg·K

WORK:

q = m·C·T

m = 32 g = 0.032 kg

T = 20°C - 60°C = – 40°C

q = (0.032kg)(-40°C)(235J/kg·K)q = – 301 J

K. Heat Transfer

How much heat is required to warm 230 g of water from 12.0°C to 90.0°C?

GIVEN:

m = 230. g

Ti = 12.0°C

Tf = 90.0°C

q = ?

C= 4184 J/kg·K

WORK:

q = m·C·T

m = 230 g = 0.23 kg

T = 90°C - 12°C = 78°C

q = (0.23kg)(78°C)(4184 J/kg·K)q = 75,000J

K. Heat Transfer A 0.400 kg sample of glass requires 3190 J for its

temperature to increase from 273 K to 308 K. What is the specific heat for this type of glass?

GIVEN:

m = .400 kg

Ti = 273 K

Tf = 308 K

q = 3190 J

C= ? J/kg·K

WORK:

q = m•C•T

T = 308 K – 273 K = 35 K

(3190 J) = (0.400 kg)(35 K)C

Kkg

JC

35400.

3190

C= 228 J/kg·K

K. Heat Transfer

How much energy would be absorbed by 550 g of copper when it is heated from 24°C to 45°C?

GIVEN:

m = 550 g

Ti = 24°C

Tf = 45°C

q = ?

C= 0.38 J/g·K

WORK:

q = m·T·Cp

T = 45°C - 24°C = 21°C

q = (550g)(21°C)(0.38 J/g·K)q = 4,400 J

CCu= 0.38 J/g·K

Assignments

HomeworkFinish Thermal Energy WS – Work on it now so you can get some help!